I have a function defined as
$f(x) = e^{-\frac{1}{x^2}}, $if $ x\ne0$; $0$ if $x =0$.
where $f:[0,\infty) \to \mathbb{R}$
I am asked to prove the following:
(a) that the nth derivative is of the form:
$f^{(n)}(x) =e^{-\frac{1}{x^2}}\frac{P_n (x)}{x^{N_n}}$
where $P_n (x)$ is a polynomial and $N_n \in \mathbb{N}$.
(b) $f^{(n)}(0) = 0$
(c) The Taylor Polynomials of $f$ around $0$ do not converge except on the point $x=0$.
For the point (a) I proceeded by induction over n, and used the fact that product and sum of polynomials is a polynomial, and so is the derivative of a polynomial.
However I am having some trouble with point (b) I tried to calculate
$\lim_{x\to 0}e^{-\frac{1}{x^2}}\frac{P_n (x)}{x^{N_n}}$
taking $x=\frac{1}{t}$ and calculating the limit as $t\to\infty$. But, because of the product rule I always have a $t^{N_n}P_n^{(n)} (x)$ and I don't know the degree of $P_n(x)$ (it gets pretty messy).
EDIT Am I a least on the right track? Or am I missing the point?
I would appreciate any hints and/or comments. Thank you for reading.
Since $(x^{-k})' =(-k)x^{-k-1} $, he first terms of $f^{(n)}(x)$ are $f'(x) =(-2x^{-3})e^{-x^{-2}} =\frac{-2}{x^{3}}e^{-x^{-2}} $ and $f''(x) =e^{-x^{-2}}(6x^{-4}+(-2x^{-3})^2) =e^{-x^{-2}}(6x^{-4}+4x^{-6}) =e^{-x^{-2}}\frac{6x^2+4}{x^6} $.
Suppose $f^{(n)}(x) =\frac{p_n(x)}{x^{cn}}e^{-x^{-2}} =p_n(x)x^{-cn}e^{-x^{-2}} $. Since $(p_n(x)x^{-cn})' =p'_n(x)x^{-cn}-cnp_n(x)x^{-cn-1} =(xp'_n(x)-cnp_n(x))x^{-cn-1} $,
$\begin{array}\\ f^{(n+1)}(x) &=(f^{(n)}(x))'\\ &=(p_n(x)x^{-cn}e^{-x^{-2}})'\\ &=(p_n(x)x^{-cn})'e^{-x^{-2}}+(p_n(x)x^{-cn})(e^{-x^{-2}})'\\ &=(p_n(x)x^{-cn})'e^{-x^{-2}}+(p_n(x)x^{-cn})(2x^{-3}e^{-x^{-2}})\\ &=e^{-x^{-2}}((p_n(x)x^{-cn})'+2p_n(x)x^{-cn-3})\\ &=e^{-x^{-2}}((xp'_n(x)-cnp_n(x))x^{-cn-1}+2p_n(x)x^{-cn-3})\\ &=e^{-x^{-2}}x^{-cn-3}((x^2(xp'_n(x)-cnp_n(x))+2p_n(x))\\ &=e^{-x^{-2}}x^{-cn-3}(x^3p'_n(x)-cnx^2p_n(x)+2p_n(x))\\ &=e^{-x^{-2}}x^{-cn-3}(x^3p'_n(x)-(cnx^2-2)p_n(x))\\ \end{array} $
If there are no errors here (prob < .7), $c=3$ works and then $p_{n+1}(x) =x^3p'_n(x)-(3nx^2-2)p_n(x) $.
Note that $p_{n+1}(0) =2p_n(0) $, so all of the $p_n(0)$ are non-zero (and, in fact, $p_n(0)=2^{n+1}$).
So $f^{(n)}(x) =p_n(x)x^{-3n}e^{-x^{-2}} $. Therefore
$\begin{array}\\ \lim_{x \to 0} f^{(n)}(x) &=\lim_{x \to 0} p_n(x)x^{-3n}e^{-x^{-2}}\\ &=p_n(0)\lim_{x \to 0} p_n(0)x^{-3n}e^{-x^{-2}}\\ &=p_n(0)\lim_{x \to 0} e^{-x^{-2}-3n\ln(x)}\\ &=p_n(0)\lim_{x \to 0} e^{-x^{-2}(1+3nx^2\ln(x))}\\ \end{array} $.
Since $\lim_{x \to 0}x \ln x =0 $ (setting $x = 1/y$, $x \ln x =(1/y)\ln(1/y) =-\ln(y^{1/y}) \to 0 $ since $y^{1/y} \to 1$ as $y \to \infty$ . Therefore
$\begin{array}\\ \lim_{x \to 0} f^{(n)}(x) &=p_n(0)\lim_{x \to 0} e^{-x^{-2}(1+3nx^2\ln(x))}\\ &=p_n(0)\lim_{x \to 0} e^{-x^{-2}}\\ &= 0\\ \end{array} $.